WITH TB (L_LEVEL,PARENT_CODE,SON_CODE,FBOMID) --要有列名AS(SELECT 0 AS L_LEVEL,FPRODUCTNO AS PARENT_CODE,FPRODUCTNO AS SON_CODE,FID AS FBOMID FROM T_BOM WHERE fproductno=‘R00127‘ AND fflag=1union allselect TB.L_LEVEL+1 AS L_LEVEL,A.PARENT_CODE,A.SON_CODE,(select MAX(fid) from t_bom where fproductno=A.SON_CODE AND fflag=1 ) AS FBOMIDfrom PRODUCT_RELATION A inner join TB on A.PARENT_CODE=TB.SON_CO...
前段时间用类似于散弹式编程的方式,各种猜测-运行验证-修正结果,最终成功转换了一个看起来比较有难度的递归函数.但总觉得很蛋疼,原因如下:1.虽然正确,但是逻辑搞得比较复杂.现在去看,一头雾水,不知道当时是怎么想的,也没有任何欲望去理解.2.没有充分利用本地变量,而是保守地把当前栈的数据一股脑绑定到Stack对象上.3.我需要用一个Class来保存栈吗?是不是太重了?为什么不考虑用一个简单的tuple来实现呢?def recur(n):if n==1:return...
<?phpheader("content-type:text/html;charset=utf8"); set_time_limit(0); $dir = "d:\\";function show ($dir){$handle = @opendir($dir);echo "<ul>";while($file = @readdir($handle)){if($file == "."||$file == "..")continue;if(is_dir("$dir/$file")){show("$dir/$file");}else{if(pathinfo("$dir/$file",PATHINFO_EXTENSION) == "pdf"){// copy("$dir/$file","./pdf/$file");echo "<li>".iconv(‘gbk‘,‘utf-8‘,$file)....
一、递归函数一般递归100多次,都没有解决的问题,放弃递归。默认递归深度:998修改默认递归次数import syssys.setrecursionlimit(100000)#设置十万次count = 0def func1():global countcount += 1print(count)func1()func1()用递归 解决一个年龄问题。alex 他比佩奇 大两岁。 4 age(3) + 2佩奇 他比日天 大两岁。 3 age(2) + 2日天 他比太白 大两岁。 2 age(1) + 2太白:我今年23. 1 23def age(n):if n == 1:...
WITH UserID (OrgCode, ParentCode, Level) AS ( SELECT OrgCode,ParentCode,0 AS Level FROM UMS_OrganizationWHERE OrgCode= ‘ORG12743‘ UNION ALL SELECT e.OrgCode,e.ParentCode, Level + 1 FROM UMS_Organization AS e INNER JOIN UserID AS d ON e.ParentCode = d.OrgCode ) SELECT OrgCode FROM UserID http://www.cnblogs.com/xqhppt/archive/2011/02/15/1955366.htmlSQL语句-递归标签:本文系统来源:http://...
公用表达式的定义非常简单,只包含三部分: 公用表表达式的名字(在WITH之后) 所涉及的列名(可选) 一个SELECT语句(紧跟AS之后) 在MSDN中的原型:WITH expression_name [ ( column_name [,...n] ) ] AS ( CTE_query_definition ) 按照是否递归,可以将公用表(CTE)表达式分为递归公用表表达式和非递归公用表表达式. 非递归公用表表达式(CTE) 非递归公用表表达式(CTE)是查询结果仅仅一次性返回一个结果集用于...
向下查询 WITH RECURSIVE res AS ( SELECT * FROM t_tree WHERE id = 2union ALL SELECT t_tree.* FROM t_tree, res WHERE t_tree.pid = res.id ) SELECT * FROM res ORDER BY id limit 3 OFFSET (2-1)*3;--分页--向上 WITH RECURSIVE res AS ( SELECT * FROM t_tree WHERE id = 2union ALL SELECT t_tree.* FROM t_tree, res WHERE t_tree.id = res.pid --select * from t_tree where t_tree.id = (SELECT t_tree.pid FROM t...
1. 文法 G(S): (1)S -> AB (2)A ->Da|ε (3)B -> cC (4)C -> aADC |ε (5)D -> b|ε 验证文法 G(S)是不是 LL(1)文法? 解:Select(A -> Da) = First(Da) = {b,a}Select(A -> ε) = (Follow(ε)-{ε})∪Follow(A) = {b,a,c,ε}Select(C -> aADC) = First(aADC) = {a}Select(C -> ε) = (Follow(ε)-{ε})∪Follow(C) = {ε}Select(D -> b) = First(b) = {b}Select(D -> ε) = (Follow(ε)-{ε})∪Follow(D) = {a,ε...
1. 文法 G(S): (1)S -> AB (2)A ->Da|ε (3)B -> cC (4)C -> aADC |ε (5)D -> b|ε 验证文法 G(S)是不是 LL(1)文法? FIRST集 FIRST(Da)={b,a} FIRST(ε)={ε} FIRST(cC)={c} FIRST(aADC)={a} FIRST(b)={b} FOLLOW集 FOLLOW(A)={c,b,a,#} 其中#是FOLLOW(A)=FOLLOW(C)=FOLLOW(B)=FOLLOW(S) FOLLOW(B)={#} FOLLOW(C)={#} FOLLOW(D={a,#} SELECT集 SELECT(A->Da)=FIRST(Da)={b,a} SELECT(A->ε)=FOLLOW(A)={c,...
原文链接:http://www.cnblogs.com/linlin/archive/2011/04/20/2021863.html首先表结果如下select MID,MPID from Member_Tbl 比如要找到MID=1的所有子级元素,子级元素在找子子级元素...,直至null,也就是向下递归declare @MID intset @MID=1;with cte as(select MID,MPID from Member_Tbl where MID=@MIDunion allselect m.MID,m.MPID from cte inner join Member_TBl m on cte.MID=m.MPID)select * from cte结果:如果要从子级元...